   Chapter 10.4, Problem 13E

Chapter
Section
Textbook Problem

# Graph the curve and find the area that it encloses.13. r = 2 + sin 4θ

To determine

To find: The area of the region that the polar equation encloses.

Explanation

Given:

The polar equation is r=2+sin4θ .

Assume the value of angle θ=180° .

Calculate the value of r.

r=2+sin4θ=2+sin4(180×π180)=2

Calculate the value of x coordinate of curve.

x=rcosθ

Substitute 2 for r and 180° for θ .

x=rcosθ=2cos(180×π180)=2

Calculate the value of y coordinate of curve.

y=rsinθ

Substitute 2 for r and 180° for θ .

y=2sin(180×π180)=0

Repeat the calculation of the values of x and y using the value of θ from 180° to 180° .

Tabulate the values of x and y in Table (1).

 θ r=2+sin4θ x=rcosθ y=rsinθ −180 2 −2 0 −170 2.642788 −2.60264 −0.45892 −160 2.984808 −2.8048 −1.02086 −150 2.866025 −2.48205 −1.43301 −140 2.34202 −1.79409 −1.50542 −130 1.65798 −1.06573 −1.27009 −120 1.133975 −0.56699 −0.98205 −110 1.015192 −0.34722 −0.95397 −100 1.357212 −0.23568 −1.33659 −90 2 1.23E-16 −2 −80 2.642788 0.458915 −2.60264 −70 2.984808 1.020864 −2.8048 −60 2.866025 1.433013 −2.48205 −50 2.34202 1.505422 −1.79409 −40 1.65798 1.270086 −1.06573 −30 1.133975 0.982051 −0.56699 −20 1.015192 0.953969 −0.34722 −10 1.357212 1.336593 −0.23568 0 2 2 0 10 2.642788 2.602638 0.458915 20 2.984808 2.804802 1.02084 30 2.866025 2.482051 1.433013 40 2.34202 1.794092 1.505422 50 1.65798 1.065729 1.270086 60 1.133975 0.566987 0.982051 70 1.015192 0.347216 0.953969 80 1.357212 0.235677 1.336593 90 2 1.23E-16 2 100 2.642788 −0.45892 2.602638 110 2.984808 −1.02086 2.804802 120 2.866025 −1.43301 2.482051 130 2.34202 −1.50542 1.794092 140 1.65798 −1.27009 1.065729 150 1.133975 −0.98205 0.566987 160 1.015192 −0.95397 0.347216 170 1.357212 −1.33659 0.235677 180 2 −2 0

Plot the graph as below.

From the graph, it is observed that the curve lies on the entire quadrant. Thus, the limit tends to be between 0θ2π .

Calculate the area of the region using the polar area formula

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